If $ a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = 1$, then show $ab+bc+ca> \frac{-1}{2}$

Question

If $ a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = 1$, then show that $ab+bc+ca\ge \frac{-1}{2}$

If figured out that if I put $(a+b+c)^2 = 0$ then I will get the above answer, but $(a+b+c)^2 = 0$ is not given in the question, so is there any other method to do it, or the question is wrong?

Answer 1

You should use $0\le(a+b+c)^2=1+2(ab+bc+ca)$. Your inequality actually shouldn't be strict, because e.g. we can take $a=0,\,b=-c=\frac{1}{\sqrt{2}}$ to get $a+b+c=0,\,a^2+b^2+c^2=1$.